You have a number of envelopes with widths and heights given as a pair of integers (w, h). One envelope can fit into another if and only if both the width and height of one envelope is greater than the width and height of the other envelope.
What is the maximum number of envelopes can you Russian doll? (put one inside other)
Example:
Given envelopes = [[5,4],[6,4],[6,7],[2,3]], the maximum number of envelopes you can Russian doll is 3 ([2,3] => [5,4] => [6,7]).
一刷
題解：
首先對數組根據width進行排序, 如果width相同, height遞減
然后就變成了尋找longest increasing subsequence的子問題。注意，這個dp數組存儲遞增序列。
Time complexity O(nlogn), space complexity O(n)
public class Solution {
public int maxEnvelopes(int[][] envelopes) {
if(envelopes == null || envelopes.length == 0) return 0;
//Ascend on width and descend on height if width are same.
Arrays.sort(envelopes, new Comparator<int[]>(){
public int compare(int[] a, int[] b){
if(a[0] == b[0]) return b[1] - a[1];
else return a[0] - b[0];//first compare width
}
});
int[] dp = new int[envelopes.length];
int len = 0;
for(int[] envelope : envelopes){
int index = Arrays.binarySearch(dp, 0, len, envelope[1]);
if(index<0) index = -(index+1);//the insert point
dp[index] = envelope[1];//height
if(index == len) len++;
}
return len;
}
}
二刷：
注意，為了不把[4,5], [4,6]算做一組有效的值，當長相等時，寬用逆序。保證只有一個被算進去。
排序后變為[4,6],[4,5].
分析：怎樣用binarySearch找到需要insert的地方。用最典型的binarysearch的寫法就能實現。因為，最后一個點的時候，i,j肯定相鄰，mid處于i的地方，如果大于i，則Insert的地方為j， 否則為i. 即滿足lo == hi的點。但是注意dp的數組需要先fill Integer.MAX_VALUE
public class Solution {
public int maxEnvelopes(int[][] envelopes) {
int len = envelopes.length;
Arrays.sort(envelopes, new Comparator<int[]>(){
public int compare(int[] a, int[] b){
if(a[0] == b[0]) return b[1] - a[1];
else return a[0] - b[0];
}
});
int[] dp = new int[len];
Arrays.fill(dp, Integer.MAX_VALUE);
len = 0;
for(int[] env : envelopes){
int index = bs(dp, 0, len, env[1]);
dp[index] = env[1];
if(index == len) len++;
}
return len;
}
public int bs(int[] dp, int lo, int hi, int target){
while(lo<=hi){
int mid = lo + (hi-lo)/2;
if(dp[mid] == target) return mid;
else if(dp[mid]<target) lo = mid+1;
else hi = mid-1;
}
return lo;
}
}
三刷
題解：有兩個需要特別注意的地方
一個是sort的時候，如果根據width排序，如果width相等，height高的在前。
一個是dp數組初始化的時候要Arrays.fill(dp, Integer.MAX_VALUE), 這樣才能用binarySearch找到正確的位置。
class Solution {
public int maxEnvelopes(int[][] envelopes) {
Arrays.sort(envelopes, new Comparator<int[]>(){
public int compare(int[] a, int[] b){
if(a[0] == b[0]) return b[1] - a[1];
else return a[0] - b[0];
}
});
//longest ascending subsequence
int len = 0;
int[] res = new int[envelopes.length];
Arrays.fill(res, Integer.MAX_VALUE);
for(int i=0; i<envelopes.length; i++){
int height = envelopes[i][1];
int index = Arrays.binarySearch(res, height);
if(index<0) index = -(index+1);
if(index == len) len++;
res[index] = height;
}
return len;
}
}
